1 Answer
1

First of all, the integrand is symmetric with respect to $r=t/2$, so that
$$
h(t)=2(1+t)^d\int_0^{t/2} (1+t-r)^{-d} (1+r)^{-d}\,dr.
$$
The function $(1+t-r)(1+r)$ is increasing and concave on the interval $[0,t/2]$. It follows that
$$
(1+t-r)(1+r)\ge1+t+\frac t2\,r,\quad 0\le r\le t/2.
$$
Then
$$
h(t)\le2(1+t)^d\int_0^{t/2} \Bigl(1+t+\frac t2\,r\Bigr)^{-d}\,dr.
$$
It follows immediately that $h(t)\le t$ for all $t\ge0$. On he other hand, evaluating the integral we get
$$
h(t)\le\frac{4(1+t)^d}{(d-1)t}\Bigl((1+t)^{-d+1}-(1+t+\frac{t^2}{4})^{-d+1}\Bigr)\le\frac{4(1+t)}{(d-1)t}.
$$
Thus
$$
h(t)\le\max\Bigl(t,\frac{4(1+t)}{(d-1)t}\Bigr),\quad t\ge0.
$$
Evaluating at the point in which
$$
t=\frac{4(1+t)}{(d-1)t}
$$
we finally get
$$
h(t)\le\frac{2(1+\sqrt d)}{d-1},\quad t\ge0.
$$